How could D.C. draw correct conclusions from just his “vision” (?) of dynamical phenomena, if:A) No matter how many times he is told not rectilinear movements ALWAYS require some force perpendicular to the trajectory to intervene, with the FUNCTION of centripetal force, he despises that basic Physics fact …B) He insists in calling a revolving movement around moon-earth barycenter just a movement “perpendicular” to straight line gravitational forces, even having been told that it is a two-dimension movement, and that the acceleration inherent in that type of movements (centripetal) is perpendicular to the speed of the revolving (or rotating) object, therefore with same direction of gravitational forces in our case. And that, logically, also inherent inertial force (centrifugal) has same direction...C) He seems to ignore the concept of “inertia" … or at least never mentions it, or “inertial” as an adjective applied to concepts such as “force”, “effect” … D) ... ("impossible" to list all his misconceptions ...) BY THE WAY, on a link posted by PmbPhy some days ago it is said:"… the production of ocean tides is basically the CONSEQUENCE of the gravitational action of the moon- and, to a lesser extent, the sun …The analysis of the phenomenon is, however, considerably helped by introducing the concept of INERTIAL FORCES as developed in the present chapter.… With respect to the CM of the earth-moon system … the earth´s center of mass has an acceleration of magnitude Ac … every point in the earth receives this same acceleration from the moon´s attraction.… for a particle at the earth´s center, the CENTRIFUGAL FORCE and the moon´s gravitational attraction are equal and opposite. If, however, we consider a particle on the earth´s surface at the nearest point to the moon …, the gravitational force on it is greater than the CENTRIFUGAL FORCE by an amount that we shall call Fo …By an exactly similar calculation, we find that the tide-producing force on a particle of mass mat the farthest point from the moon … is equal to -Fo …”What I´ve been saying here time and again !!LOGICALLY, the authors, as the NOAA scientists D.C. loftly despises (!!), neither say “the production of ocean tides is ONLY due to differential gravitational action" (or something with that meaning), nor “forget” an acting inertial force such as centrifugal force, as he stubbornly keeps doing !!

Show your numbers then and claim your Nobel Prize if they vary depending on the direction and speeds of planet and moon at moments when the two are the same distance apart. The perpendicular component of motion does not affect the tidal forces. All you've done is make an artificial divide by splitting the gravitational force into centripetal and non-centripetal, and once you've added everything up, you'll be left with the same forces that are worked out much more simply just by treating the gravitational force as a single force (which is exactly what it is).

As for not mentioning inertia, it should be obvious to anyone that there is such a thing, but with gravitational pull, the speed of acceleration of an object being accelerated by gravity is the same regardless of the mass involved, so it has no impact on how much the Earth and moon accelerate towards each other (one ton of rock is accelerated just as much as a billion tons of rock ). The further parts are pulled less strongly and the near parts more strongly, so forces are generated as those parts try to lift. That's all there is to it and you need to stop asserting otherwise.

What an "interesting" case for the study Neil Comins did:Sources of Misconceptions in Astronomy, by Neil F. CominsUniversity of Maine

"...In pursuing them and their origins, I use the definition of misconception as a resilient concept held to be correct, but that is actually at variance with accepted scientific knowledge. Resilient here means that the concept is incorporated in a person's conceptual framework and therefore resistant to change. Because they are woven into conceptual fabrics, many misconceptions help prevent further correct understanding of related topics. The description of "conceptual barriers to learning," by Hawkins, Apelman, and Colton & Flexner, cited in Confrey (1987) helped me focus on an operating definition.For the past eighteen months I have been working with students taking the above- mentioned introductory college astronomy course in an effort to understand the origins of their misconceptions about astronomy. I will describe the methodology I am using shortly. In the process, the students have shared with me over five hundred, fifty misconceptions they held prior to taking the course. In this paper I report on a set of origins of astronomical misconceptions that derives from their lists and from interviews I held with small groups of students. The origins, listed below, can be used to explain virtually all the stated misconceptions. This set is certainly not unique; other categories of internal and external error leading to misconceptions can certainly be identified or constructed. Nor is this set necessarily complete, although it has been satisfactory in explaining the misconceptions on hand, a list that is continually growing. It is also important to point out that the list of origins is heterogeneous, some deriving from sources external to the holder of the misconception, some internal, and some with both internal and external characteristics.

I can't follow it well enough to be able to comment on it in any useful way, and the same applies to the new stuff. You seem to be introducing complications which obscure things rather than simplifying things to isolate the different factors.

I guess that what you can't follow is not the way we can change the trajectory of satellites. It is well established that if we accelerate a satellite that is already on a circular orbit, it will start decelerating and getting away from the earth until it gets at is apogee, and vice-versa if we decelerate it. Thus what you can't do is probably applying this principle to the tides, or to any example where different parts of an object are not traveling at the same orbital speed as its center of gravity. Instead, you prefer making your point with your own examples. I think you don't need to do that with me, because I already understand your point. Moreover, my principle doesn't seem to contradict yours: whenever we stop the orbital speed in my example, the tidal bulges stay the same, because they are not affected by the rotational inertial force as in rmolnav's principle. Nevertheless, inertia is certainly accounted for when we accelerate or decelerate a satellite, and the change in speed certainly changes the parameters of its orbital trajectory with time, so applying that principle to tides doesn't seem to contradict rmolnav's principle either.

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You seem to be introducing complications which obscure things rather than simplifying things to isolate the different factors.

Your principle is simple for bodies with no orbital speed, but it is certainly not that simple for those with speed otherwise I guess rmolnav would have accepted it. I do accept it, but I had to be able to figure out how inertial motion would affect it. In your simulation, I assume that the whole earth (with its bulges) is first moved tangentially away from the moon a bit, and that you then apply the gravitational force at that distance to bring it back on its trajectory. The first motion is thus inertial, and the second one is gravitational. This way, there is visibly no need to use any inertial force to take the earth away, it moves all by itself and the bulges too, and when the gravitational force begins to pull, it pulls differently on its different parts, so the bulges change if the distance to the moon has changed. It works fine, and my explanation also works fine since the orbital speed I'm using to move the bulges tangentially away would be simulated the same way. What would be different is that the tangential orbital speed of the earth surface would change constantly due to its daily rotation, and as I explained, that should decelerate its rotation with time.

Resilient here means that the concept is incorporated in a person's conceptual framework and therefore resistant to change.

We all resist to change, and every existing thing does, so it is useless to try to bypass the problem. The only way is to deal with it, and as far as our ideas are concerned, it is to go on pushing them until others agree with them, but there is still a bug, because people can visibly adopt false ideas.

Nevertheless, inertia is certainly accounted for when we accelerate or decelerate a satellite, and the change in speed certainly changes the parameters of its orbital trajectory with time, so applying that principle to tides doesn't seem to contradict rmolnav's principle either.

If you look at the perpendicular component of movement, there is no acceleration (or deceleration) - the inertia of the bodies just keeps them moving that way at a constant speed (until there's a change in the direction the force is being applied in). The only change occurs along the direction in which the force is acting, and on that line you get the same size of tidal forces being generated regardless of whether the bodies are moving apart or together at that time. The perpendicular component is irrelevant.

As I said before. rmolnav needs to show his numbers if he thinks otherwise. I wrote a program which shows my numbers, and I invited him to produce his maths to put into the program to replace mine. He hasn't done so, and he probably isn't even capable of doing so. If he ever does though, I will be more than happy to take his code of this (pseudo-code will do - the algorithm is all I need from him because it isn't my job to build that for him) and put it into the program for him, and it can then run both methods at the same time while displaying the numbers for each side by side. If his maths is right, the numbers will necessarily match up with mine, but it will take a lot more processing to work them out each time because he is splitting the task up into artificial components. There is only one force driving tidal forces, and that's gravitational pull acting in a straight line and reducing in strength over distance.

Think about what happens when the Earth is directly between the sun and moon. What is the mechanism now for the tidal forces? Let's move the Earth and moon to a different orbit to make the two lots of pull equally strong. Be aware that this means moving them further away from the sun rather than nearer to it - the sun's input to the tidal forces is only smaller because its pull varies less across the Earth than the moon's pull does due to the greater distance to the sun. The Sun is pulling more strongly on F, the far side from the moon, and less strongly on N, the near side to the moon, while the moon is pulling more strongly on N and less strongly on F. These pulls are in opposite directions, so the result is enhanced tidal forces. That is the correct explanation. But what happens if we start chopping up the forces in the way rmolnav seeks to do and attributing extra mechanisms to the system?

If we're playing rmolnav's game, we take each of these forces at the centre of the Earth and label them as centripetal force. At point F (the far side of the Earth from the moon), we attribute the material's trying to lift there mainly to centrifugal force, plus a little bit to the sun's greater pull there. At point N (the near side of the Earth to the moon), we attribute the material's trying to lift there mainly to the moon's greater pull, plus a little due to centrifugal force. We supposedly have centrifugal force trying to fling material out at opposite sides in opposite directions. The Earth is neither accelerating towards the sun or moon in this situation (because we've put the Earth at a distance from the sun where the two forces are equally strong), so the Earth is moving in a straight line (and is doing so in every frame of reference). Let me spell that out more clearly for rmolnav: the Earth is moving in a straight line and you think you have material being flung out sideways by centrifugal force. No - what's actually happening is that you're mauling the physics. Everything that's actually happening is down to straight-line differential gravity.

Let's do another version which makes things even more stark. Let's give the Earth two moons which always stay on opposite sides. The Earth isn't moved by them at all now, but we still have tidal forces, nearly twice as strong as we have with just the one moon. How does rmolnav's explanation work here? The gravitational pull from each at the Earth's centre is called centripetal force, and the two lots cancel out. Each bulge is driven partly by centrifugal force from the Earth's "movement" acting on the material furthest from the moon that's applying the centripetal force to the centre that drives this "movement", and partly by the greater gravitational pull applied there by the other moon. Take away the sun and the Earth's orbit so that all we have is the Earth and its pair of moons, and what do we have this time? We have a stationary Earth with material being flung out at opposite sides by, well, I suppose it should be called centrifugal farce.

Ask yourself now, who is the person here who repeatedly fails to understand the basic physics that's at work here? Is it the one who can't supply his numbers and who keeps arguing against the correct explanation while pushing a much more complex and contrived one where a force is artificially split into different components and where that explanation breaks down in multiple situations, or is it the one who supplies actual numbers, properly reasoned arguments backed by a computer simulation which runs his model and demonstrably works, and whose explanation covers all cases?

In your simulation, I assume that the whole earth (with its bulges) is first moved tangentially away from the moon a bit, and that you then apply the gravitational force at that distance to bring it back on its trajectory. The first motion is thus inertial, and the second one is gravitational. This way, there is visibly no need to use any inertial force to take the earth away, it moves all by itself and the bulges too, and when the gravitational force begins to pull, it pulls differently on its different parts, so the bulges change if the distance to the moon has changed.

If I am right about the way your simulation works, it is so simple that I think Rmolnav should be able to understand it if we describe it properly. It contains the two principles, inertial and gravitational, and no inertial force is needed. I think the misunderstanding comes from using force instead of motion to describe the tangential motion. if we throw a stone with a sling, it is not a force that moves the stone away from us, only the tangential motion. The stone doesn't move radially away from us, it moves tangentially away from the point where it was released by the rotating sling, it transforms the rotational speed of the sling into inertial motion. Of course that the pulling from the moon exerts a force on the earth's mass so that it doesn't get away, but we can easily see that it is not a force that moves the earth away from the moon either, only a tangential motion. That straight tangential motion could also be curved by the pulling of a sling if there was no gravitation, but in this case, only the pulling of gravitation is at work.

David, you did not comment a part of my previous answer that I found interesting

I didn't have anything to say about it and still don't. If something's moving perpendicular to the force that's being applied, it simply maintains that perpendicular component of the movement while the acceleration adds or reduces movement in the direction of the force. After that change has been made, the object will naturally be moving in a slightly different direction from the previous one (unless the perpendicular component was zero). The result is simply what comes from that, and there's no other factor, except that any non-zero perpendicular component of movement leads to the next application of the force being applied in a slightly different direction the next time. Where material goes or tries to go will fit in perfectly with the idea of it being directed there by "centrifugal force" in some cases, but it's actually just going where it goes (or trying to) based on the amount of gravitational force acting on it and trying to alter its current movement accordingly.

We can see now a clear case in my previous post involving a planet with a moon going round it where rmolnav's explanation breaks horribly, and I want to expand that a bit further.

I did a few in-the-air calculations last night and reckon that if you put the Earth and moon about ten times as far from the sun as they actually are (i.e. to somewhere near Saturn's orbit), the strength of the Sun and moon's gravitational pull on the Earth would be equal. [If someone else wants to attempt that calculation, it would be interesting to see if we agree, but the principle involved here is right regardless of the actual distance required.] Imagine the Earth between the moon and the sun there. The pull is equal (on average) from each, so the Earth must be moving in a straight line at that moment. The strength of the moon's gravitational pull on the near and far sides of the Earth varies more than the Sun's because the moon and Earth are close together with the moon's gravity falling off in strength quickly over distance due to the proximity of the source. The sun also drives much smaller tidal forces because the strength of its gravity falls less slowly over distance by the time it reaches the Earth, and as we're putting the Earth out near Saturn, that fall off in strength over distance will be ten times smaller still, meaning that it won't add much to the tides at all.

Rmolnav's explanation says that the the bump on the far side from the moon is caused by centrifugal force, but in this case it clearly can't be because the Earth's moving in a straight line. The bump is still there though in full (and with a tiny bit extra added by the sun). So in this situation, what does he have to do to account for that bulge? He has no option other than to switch to the straight-line differential gravity explanation. He can justify this though on the basis that the sun and moon's pull cancel out all the centripetal force and leave the gravitational force simply being counted as one thing instead of being (artificially) divided. Take away the sun, and then he says the same tidal force (minus the tiny bit that was added by the sun) is now suddenly caused by centrifugal force instead.

The Earth obviously isn't as far out as Saturn, so let's now look at it where it actually is. We have situations once a month here where the Earth is between the moon and the sun, and it isn't following a straight line, but a curve that curves the opposite way from the one rmolnav normally thinks of it doing. If he wants to have centrifugal force involved here, it has to be throwing a "bulge" towards the moon rather than away from it, and yet it isn't the sun's gravity that's driving most of that "bulge" - it's the moon's stronger pull there that's responsible - the only component of the tidal force there that could be attributed to centrifugal force is the part caused there by the sun. Think about that carefully - this is a more severe curve that the Earth's following than the one in a case where the sun doesn't exist, and it curves the opposite way too. That should mean much greater centrifugal force, but at the same time, this is balanced out by the sun's stronger gravitational pull on the material that's trying to lift, and by the fact that the strength of the sun's pull falls off less over distance at this range. The end result is that the part of the "bulge" here (on the side of the Earth nearest the moon) which is attributed to centrifugal force would indeed only be the part caused by the sun, while the larger component of the "bulge" is attributed to the moon's differential gravity even though the moon is on the outside of the curve of the path that the Earth is following, and that component of the "bulge" is not driven by centrifugal force in any way. This is quite amusing - we have two "bulges" in the same place (and adding together into a bigger "bulge"), one caused by centrifugal force and the other not caused by centrifugal force at all. Here it becomes manifest just how ludicrous rmolnav's explanation actually is.

(None of the bulge on the side nearest the sun is being driven by centrifugal force because it's at the inside of the curve, and most of it isn't being lifted by the sun's differential gravity, so what is lifting it? It is actually being pulled up by the sun's gravity, but it's primarily the moon's differential gravity holding it back less strongly there than the rest of the Earth that allows it to try to lift.)

The numbers do actually match up if you crunch them all, but this only happens as a necessary accident. The reality is that this business of dividing gravity up into centripetal and non-centripetal force is just a silly abstraction which necessarily fits the numbers, but has no mechanistic role in what's happening whatsoever. The real mechanism is simply straight-line differential gravity in all cases. The real mechanism is the simple addition of all the gravitational forces acting on each particle - you sum the vectors and end up with one single result per particle which dictates how it tries to accelerate. The physics of this is ridiculously simple, but when people learn too much through abstractions, they can be misled by their education and end up tripping up over issues of this kind. Even good scientists can make such errors, but there is no shame attached to them from doing so - to err is human.

If we consider a case where the Earth is at the 90 degree corner of a right-angled triangle with the moon and sun at the other corners. What's going on in this case? There are four "bulges", but the smaller ones appear as "dips". Well, we know these bulges aren't real - they merely drive the generation of real tidal bulges. We actually have four sides with tidal forces acting to lift material there. If we attribute one of them to centrifugal force, it has to be the one on the far side of the Earth from the sun. The other three, including the pair caused by the moon, are driven by differential gravity and not by centrifugal force. The more you look at this, the more rmolnav's explanation falls apart. The only question remaining to be answered now is whether he recognises that.

Sorry, but rubbish and more rubbish ... !!Another day, if I manage to keep patient, I´ll comment on last posts. But now, something about another of a few days ago.On #211 I said:"...IN A QUITE SIMILAR WAY, we could reason as follows:If, somehow (a “megaman” grasping the earth at its c.g…?), we “removed” that rectilinear, accelerated movement of earth towards the moon, outer bulge would not only disappear, but even change its sign, because outer earth hemisphere would get additional internal compressive stresses (before they were tensile s.), due to moon´s pull (though smaller than tensile stresses within inner hemisphere, closer to the moon …)Therefore, if with the differential moon´s gravity maintained, outer positive bulge doesn´t exist, the moon differential gravity cannot be the cause of outer bulge !! Same way you´ve “reasoned” lot of times … Either both are right, or both are wrong …Any guess ??"Apart from what you replied then, a few days ago you said:

Let's give the Earth two moons which always stay on opposite sides. The Earth isn't moved by them at all now, but we still have tidal forces, nearly twice as strong as we have with just the one moon. How does rmolnav's explanation work here? The gravitational pull from each at the Earth's centre is called centripetal force, and the two lots cancel out. Each bulge is driven partly by centrifugal force from the Earth's "movement" acting on the material furthest from the moon that's applying the centripetal force to the centre that drives this "movement", and partly by the greater gravitational pull applied there by the other moon. Take away the sun and the Earth's orbit so that all we have is the Earth and its pair of moons, and what do we have this time? We have a stationary Earth with material being flung out at opposite sides by, well, I suppose it should be called centrifugal farce.

What a lot of absurdities ... !!I´ve said here time and again that "centripetal force" is neither an "essentially" new force, nor just a label we can give any force "at will" !! It is just a FUNCTION, which many essentially different types of forces can EXERT, if they are CAUSING the bending of the trajectory of a moving object ...Your "second" moon, together with real one, are getting what I said on #211: the earth does´t move at all, without the necessity of any "megaman" ...And what happens is what I said then, BUT TWO OPPOSITE-WAY TIMES:"... outer bulge would not only disappear, but even change its sign, because outer earth hemisphere would get additional internal compressive stresses (before they were tensile s.), due to moon´s pull (though smaller than tensile stresses within inner hemisphere, closer to the moon …)"There is no "prevailing" centrifugal force whatsoever, because the earth doesn´t move ...And neither are there bulges caused by the differential pull from either moon ...What happens is that on each earth side the increase of sea level due to the pull from closer moon, is bigger than the DECREASE of sea level caused by the farther moon !!And you even dare say:

The physics of this is ridiculously simple, but when people learn too much through abstractions, they can be misled by their education and end up tripping up over issues of this kind. Even good scientists can make such errors, but there is no shame attached to them from doing so - to err is human.

Sources of Misconceptions in Astronomy, by Neil F. CominsUniversity of Maine

"...In pursuing them and their origins, I use the definition of misconception as a resilient concept held to be correct, but that is actually at variance with accepted scientific knowledge. Resilient here means that the concept is incorporated in a person's conceptual framework and therefore resistant to change. Because they are woven into conceptual fabrics, many misconceptions help prevent further correct understanding of related topics

BY THE WAY, on a link posted by PmbPhy some days ago it is said:...… for a particle at the earth´s center, the CENTRIFUGAL FORCE and the moon´s gravitational attraction are equal and opposite.

This illustrates the danger of considering centrifugal force to be a force. If the two forces where equal and opposite, the particles at earth’s center would be moving in a straight line, but they’re not. It is accelerating due to the one and only force acting upon it here: gravity.

Yes, the tides can sometimes be explained in these term, but by eliminating the curved trajectory of Earth by putting it further out, or by putting it between two moons as David Cooper has done, the centrifugal explanation falls apart. There is still gravity, but no centrifugal force, and yet the tide remains.

If, somehow (a “megaman” grasping the earth at its c.g…?), we “removed” that rectilinear, accelerated movement of earth towards the moon, outer bulge would not only disappear, but even change its sign, because outer earth hemisphere would get additional internal compressive stresses (before they were tensile s.), due to moon´s pull (though smaller than tensile stresses within inner hemisphere, closer to the moon …)Therefore, if with the differential moon´s gravity maintained, outer positive bulge doesn´t exist, the moon differential gravity cannot be the cause of outer bulge !!

David’s examples very much did remove that rectilinear accelerated movement, and no compression of the atmosphere on the far side of Earth results.

Quote from: rmolnav

Your "second" moon, together with real one, are getting what I said on #211: the earth does´t move at all, without the necessity of any "megaman" ...And what happens is what I said then, BUT TWO OPPOSITE-WAY TIMES:"... outer bulge would not only disappear, but even change its sign, because outer earth hemisphere would get additional internal compressive stresses (before they were tensile s.), due to moon´s pull (though smaller than tensile stresses within inner hemisphere, closer to the moon …)"

There is no compressive effect on the far side due to either moon. The attraction of the far side is less than the attraction on the Earth below it. That equates to tensile stress, and these stresses from each moon add to a double magnitude tide, not cancel out to something lower than what we see with one moon.

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What happens is that on each earth side the increase of sea level due to the pull from closer moon, is bigger than the DECREASE of sea level caused by the farther moon !!

No, they’re both increases. This seems to predict a partially cancellation of tides.

I see several references to post 211:

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If, somehow, we "remove" the revolving, both earth and moon will accelerate towards each other … Though only a transient movement, bulges would remain (and increase) until the collision.

You seem to be envisioning removal of orbital speed, not the revolving of Earth and the moon. If you remove the revolving, you get two tides per month, and we would get to see what is currently the far side of the moon.

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OK. Therefore, revolving was not causing them[/i] (YOUR deduction).IN A QUITE SIMILAR WAY, we could reason as follows:If, somehow (a “megaman” grasping the earth at its c.g…?), we “removed” that rectilinear, accelerated movement of earth towards the moon, outer bulge would not only disappear, but even change its sign, because outer earth hemisphere would get additional internal compressive stresses (before they were tensile s.), due to moon´s pull (though smaller than tensile stresses within inner hemisphere, closer to the moon …)

On this I agree. If a mysterious force grabbed only the rocky part of Earth and pushed it, but not its water away from the moon, then yes, the water would naturally slosh to the side away from the unnatural acceleration caused by this selective force. But no such megaman exists. The sun does it, but it does not decline to act on the water as well. The megaman example does not reflect what might actually keep Earth from accelerating in the various examples above.

Quote from: rmolnav on 18/09/2018 06:48:27] "...In pursuing them and their origins, I use the definition of misconception as a resilient concept held to be correct, but that is actually at variance with accepted scientific knowledge. Resilient here means that the concept is incorporated in a person's conceptual framework and therefore resistant to change. Because they are woven into conceptual fabrics, many misconceptions help prevent further correct understanding of related topics

Again, no need to accuse others to be resistant, just to consider that resistance to change is mass, that it is universal, and that it applies to anything that exists, even to our ideas. On these nice words, I think I'll leave the discussion. Sometimes it is no use to insist. Time also has its word to tell.

The biggest winner in any argument of this kind is the one who is most wrong, realises it, and shifts to a new position in order to be right. Such a winner may emerge from here some day, but there are still no signs of movement - instead we see the usual human response which is to dig a deeper trench.

I've learned a lot from this thread though by being pushed into thinking about things I wouldn't have explored otherwise, so I'm grateful to rmolnav for that - he has been useful. I'm sure he hasn't wasted any of his time here though because his superior intellect will doubtless have converted everything into a gain: as Neil Comins might say, a fool finds it very hard to learn anything from a genius, but a genius can always find something to learn from a fool.

I want to share another thing that I've learned from this. I'd never stopped to think carefully before about how sun's pull damps down the tides when the sun and moon are 90 degrees apart in the sky - I've always imagined the sun to be subtracting from the tidal forces in some way, but no: they add tidal force, causing material to try to lift out just as much as they do when the sun and moon pull from the same direction or from 180 degrees apart.

Imagine the Earth with two moons 90 degrees apart, both pulling with the same strength from the same distance away. Each moon is trying to generate two "bulges", but they are not bulges. When you combine these four bulges, you get no tides, but there are strong tidal forces trying to lift material up all round the Earth. Imagine the Earth in the middle of the screen with one moon out to the left and the other down below the Earth. The combined acceleration will move the Earth down at 45 degrees towards the bottom left. The point on the Earth close to the bottom moon is called BN (bottom near). Opposite that (on the other side of the Earth) we have the point BF (bottom far). [Maybe I should have gone for top, but no time to change it all - just press on.] The point on the Earth nearest the moon out to the side is SN (side near) and the point opposite it is SF (side far).

At BN, the material is being pulled down by the bottom moon more strongly than the Earth as a whole, but it's being pulled towards the side moon with the same strength as the Earth as a whole.

At BF, the material is being pulled down by the bottom moon less strongly than the Earth as a whole, but it's being pulled towards the side moon with the same strength as the Earth as a whole.

In both of these cases, we have tidal forces trying to lift material, and the equivalent happens at points SN and SF too, so are there four "bulges" or are there none?

Let's look at a point on the Earth's surface half way round from BN to SN, this being the point that leads the Earth towards the bottom left. The force from both moons on this point is greater than on the Earth as a whole, so again the material will lift, but this point is further away from the bottom moon than BN, and further away from the side moon than SN, so it isn't going to get double lift, and the two forces are being applied at 90 degrees, so they add as vectors, reducing the magnitude of the lift further. How much lift is it? Cos 45 = 0.707, so that's roughly the strength of each component of the lift (compared with the strength at BN or SN, and as they're acting at 90 degrees to each other, we have 0.707^2 + 0.707^2 = 1, and the square root's the same: we have the same amount of lifting force there as we do at BN and SN.

We should do the same thing at two more points to check that the tides cancel out completely (ignoring the slightly smaller tidal forces on the far side due to the strength of gravity falling off less over distance on the way there from the centre because of greater distance from the source). At the point on the surface half way between BF and SF, we have a lesser pull there with the same numbers working through to a value of 1, so it's the same lift there. One more point will complete the picture: at the point half way round from SF to BN, we have a stronger pull than average downwards, and a lesser pull than average to the left, and yes - it's the same numbers again and the same lift. So, we have no tides, but we have tidal forces all round the Earth lifting material regardless.

It the case of the real Earth, moon and sun, you can imagine this working the same way, but with the sun's input being less than the moon's (it's a greater force from the sun, but with less change over distance). When they are 90 degrees apart, all the material round the equator is being lifted a microscopic amount; more strongly in line with the moon, but still lifting all the way round. When the moon and sun are pulling from the same or opposite sides, they're then lifting material more strongly if it's in line with them, but the points on the equator which see the moon and sun down at the horizon are not being lifted at all, and it is only at times like this that the tidal force at such locations goes down to zero.

(And I apologise to rmolnav for posting more absolute cobblers here, but I kinda like correct "nonsense".)

#309 HalcThank you and welcome. You seem to argue “reasonably”, and I do hope we can discuss fruitfully …You say several things I can´t agree with.But today, I´ll only tackle what I consider ESSENTIAL for any possible understanding: NOT to have completely opposite ideas regarding INERTIA, that can manifest itself in different ways, one of them as “centrifugal force", real force (or, at least, causing quite real effects ...)

This illustrates the danger of considering centrifugal force to be a force. If the two forces where equal and opposite, the particles at earth’s center would be moving in a straight line, but they’re not. It is accelerating due to the one and only force acting upon it here: gravity.

When I first entered here (more than three years ago !!), I used to approach this issue considering earth´s CM revolving “uses” 100% moon´s gravity pull there to cause the required centripetal acceleration for that movement. No “spare” pull (which, if existing, could produce other effects) there …But on closer to moon hemisphere there is an excess of moon´s pull. Any given material particle there can´t “use” that excess of pull to get a bigger acceleration, because the rest of the planet “forces” them to move together, with same acceleration … That imbalance generates internal stresses in both senses (between contiguous parts opposite forces are exerted on each other, due to Newton´s 3rd Motion Law).The opposite occurs on farther hemisphere …THAT FIELD of interaction forces, acting either “moonwards”, or outwards (arguably “centrifugally”), would be what actually causes tides !!But I discussed the issue (directly through emails) with NOAA scientist author (at least one of them) of:https://tidesandcurrents.noaa.gov/restles3.htmland among other things he told me:"The publication you are referring to is "Our Restless Tides", a 10-page pamphlet developed in the 1950's to provide a basic description of the forces which create the tides. It's intended audience were the grade school children and adults of that time. It used terminology of science and forces which were common in the 1950s. Such as centrifugal force. Centrifugal force was always an "imaginary force" (not a real / measurable force). But that type of description made the concepts easier to understand and explain. That description and use of centrifugal force continued to be common practice until the 1970-80's. At that point, the terminology shifted and the textbooks used in grade schools were changed to use a more modern terminology and description of this "effect" being a result of inertia rather than an "imaginary force”.Initially I didn´t agree, and insisted: only previously mentioned outward internal forces could be considered “centrifugal”, to fully satisfy Newton´s 2nd Motion Law …Later, after father ruminating on the issue, certainly helped by the effort to refute many people´s propositions (here and on other sites), I got pretty sure NOAA mentioned scientists (and the authors of “French Tides” site referred to on # 276 ) are quite correct in their approach (though certainly many current books, even well known dictionaries, keep the “old” definition as an “imaginary” or “fictitious force" ...)In any case, both ways are equivalent, because deducting required centripetal force vector from moon´s pull on a given earth particle (the force the particle “interchanges” with the rest of the planet), gives same result than adding the “arguably” called centrifugal force vector to moon´s pull …But it is necessary to tackle the “apparently” insolvable fact that earth´s CM is actually accelerating … It is rather tough to explain through posts here. I did have in mind a couple of ways. With your post, yet another. In chronological order:A) One in relation to the erroneously (to me) used concept of “non-ínertial reference system" …B) Another way (though kind of equivalent) in relation to the scenarios used by D.C., imagining earth not revolving …C) Last one, in relation to what you say: “… If the two forces (moon´s pull and centrifugal force) where equal and opposite, the particles at earth’s center would be moving in a straight line, but they’re not". ANOTHER DAY I´ll try and do it.Anyhow I´ll copy and paste something I´ve already said a couple of times (way A), needing to elaborate on).PmbPhy had said:"In particular, the centrifugal force is an inertial force which is observed in rotating (non-inertial) frames”.I replied:"I´m afraid that is rather confusing ... What do you mean with "... "which is observed in …"?Does it mean it is a real force "observed" by us? Or, in that "particular" case, is it rather a kind of tricky tool WE apply to bring a "fictitious" situation (an accelerating frame of reference) back to reality (with all REAL effects which have disappeared in that frame of reference)?"

Yes, the tides can sometimes be explained in these term, but by eliminating the curved trajectory of Earth by putting it further out, or by putting it between two moons as David Cooper has done, the centrifugal explanation falls apart. There is still gravity, but no centrifugal force, and yet the tide remains.

Come on !!. You yourself are saying "by eliminating the curved trajectory" ... !!BASIC LOGICS: The fact that there can be imagined (or even real) scenarios without the intervention of any centrifugal force (logically, without any curved trajectory) where the two bulges do build, DOESN´T MEAN in other real scenarios (with curved trajectories) CENTRIFUGAL FORCE can´t be an acting REAL force …There are different ways INERTIA can manifest itself !!

BASIC LOGICS: The fact that there can be imagined (or even real) scenarios without the intervention of any centrifugal force (logically, without any curved trajectory) where the two bulges do build, DOESN´T MEAN in other real scenarios (with curved trajectories) CENTRIFUGAL FORCE can´t be an acting REAL force …

The actual case gives us situations where the curve that the Earth is following curves the opposite way to the path you claim for it. The Earth would fall 85 miles towards the moon in a day if the perpendicular component of its movement was removed, but it would also fall 14,000 miles towards the sun in a day if the perpendicular movement of its orbit round the sun was removed. Now think about the two curves added together (with the perpendicular movement not removed) and how the 85 miles one way is out-gunned by the 14,000 the other way. In this case, any centrifugal force involved would need to be trying to lift material towards the moon and not towards the sun, so the majority of the tidal force making material stick out towards the sun must come from the moon's differential gravity operating on that far side of the Earth (far side from the moon) and not from centrifugal force, while the minority of the tidal force acting to lift material there is caused by the sun's differential gravity. On the near side of the Earth (to the moon), the minority of the tidal force trying to lift material there is caused by the sun's differential gravity, and this is the only part that you could claim is caused by centrifugal force. The majority of the tidal force trying to lift material there is caused by the moon's differential gravity there, and you cannot justify calling that centrifugal force as you only have enough mathematical justification to cover the minority part of the lift there. Then when you see that the minority part of it there (the sun's contribution) is supposedly driven by centrifugal force while the majority part of it (the moon's contribution) cannot be driven by centrifugal force in any way, the mistake in your analysis becomes clear - you are making artificial distinctions. This is a real case that happens once every month.

Another real case which destroys your explanation happens twice a month, and that's the one where the sun and moon are nearly 90 degrees apart in the sky with the sun dictating most of the curved path that the Earth is following and with the moon's pull acting at 90 degrees to the direction of the only component of force that could be labelled as centrifugal. The two tidal inputs from the moon in this case can only be explained through differential gravity with no possible role for centrifugal force to aid them. These are real cases where your explanation is shown to be plain wrong, but the differential gravity explanation works in 100% of cases. An explanation which works some of the time but is wrong the rest of the time is simply an incorrect explanation. I've shown you specific cases where it's clearly 100% wrong, but there are lots of other cases with different angles where it's clearly partly wrong because the angle in which the part of the force that could be labelled as centrifugal force is not aligned with the moon. The only case where it is aligned correctly is where the moon is directly between the Earth and the sun, so there is only one moment per month when your numbers stack up 100%. The rest of the time your explanation doesn't fit the facts - it only looked as if it might fit in cases where the sun was ignored, so it turns out that you've been dealing with imaginary scenarios all this time yourself!

There is no compressive effect on the far side due to either moon. The attraction of the far side is less than the attraction on the Earth below it. That equates to tensile stress, and these stresses from each moon add to a double magnitude tide, not cancel out to something lower than what we see with one moon.

Sorry, but that is erroneous …If we want not to mix causes and effects, we must analyze the dynamics of the two-moon case very, very carefully.As a triple object and symetric system, the two moon can be rotating around the earth CM, logically at same distance and with same angular speed, and the earth stand still (for the sake of simplification, I suppose the system without any orbital movement around the sun).There are “infinite” pulls from each moon on earth particles …But the total force exerted on the earth is null …The total force exerted by each moon is equal, but opposite, to the one exerted by the other …The distribution of each moon´s pull is logically the closer, the bigger (inversely proportional to the square of the distance), logically ALWAYS towards the respective moon.Being earth not moving at all, there are no inertial effects. If now we consider SEPARATELY the effects of each moon´s pull, whatever the reason why the earth doesn´t move, each moon pulls ocean water toward itself, same way you admitted yesterday:

QuoteOK. Therefore, revolving was not causing them[/i] (YOUR deduction).IN A QUITE SIMILAR WAY, we could reason as follows:If, somehow (a “megaman” grasping the earth at its c.g…?), we “removed” that rectilinear, accelerated movement of earth towards the moon, outer bulge would not only disappear, but even change its sign, because outer earth hemisphere would get additional internal compressive stresses (before they were tensile s.), due to moon´s pull (though smaller than tensile stresses within inner hemisphere, closer to the moon …)On this I agree. If a mysterious force grabbed only the rocky part of Earth and pushed it, but not its water away from the moon, then yes, the water would naturally slosh to the side away from the unnatural acceleration caused by this selective force. But no such megaman exists. The sun does it, but it does not decline to act on the water as well. The megaman example does not reflect what might actually keep Earth from accelerating in the various examples above.

The specular feature of the whole system makes what in my analogy the "megaman" did: not to let the earth move. Further-to-each-moon hemisphere is pulled by that moon towards earth´s middle section: solid earth would be compressed (again: considering ONLY each moon effect separately), and water would move towards considered moon, causing there a sea level decrease …But closer-to-each-moon hemisphere is ALSO pulled by the moon at that side, more strongly than previously mentioned opposite pulls, causing bigger similar effects but in the opposite sense. Adding up those opposite effects (due to forces actually acting on each hemisphere particle), we have the two bulges …BUT THEY HAVE NOTHING TO DO with the so called “differential gravity” exerted separately by either moon …In all D.C. imaginary cases, in order to have a second proper bulge we must let the earth accelerate … because INERTIAL effects are absolutely necessary. So called "differential gravity”, without those inertial effects, can´t cause the second bulge …By the way, in the two-moon case the size of the bulges doesn´t increase to double whatsoever. …The second moon pull kind of substitutes real case inertial effects (centrifugal forces). To double the size of the bulges it´d be necessary to have the earth revolving around each moon-earth pair center of mass, and that is not possible simultaneously !!

And thank you for the welcome. I’ll be glad when I become human and I get more than a postage-stamp size window in which to compose my responses. As it is, I am forced to do it in an offline wordprocessor.

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But today, I´ll only tackle what I consider ESSENTIAL for any possible understanding: NOT to have completely opposite ideas regarding INERTIA, that can manifest itself in different ways, one of them as “centrifugal force", real force (or, at least, causing quite real effects ...)

I suppose it is a matter of interpretation if it is real or not. I kind of go with force being the F=ma in Newton’s 2nd law, whereas inertial effects (centrifugal being among them) are the ‘m’ in that equation, not the F. It isn’t causing any real acceleration of anything in the direction of the supposed force.

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When I first entered here (more than three years ago !!), I used to approach this issue considering earth´s CM revolving “uses” 100% moon´s gravity pull there to cause the required centripetal acceleration for that movement. No “spare” pull (which, if existing, could produce other effects) there …But on closer to moon hemisphere there is an excess of moon´s pull. Any given material particle there can´t “use” that excess of pull to get a bigger acceleration, because the rest of the planet “forces” them to move together, with same acceleration … That imbalance generates internal stresses in both senses (between contiguous parts opposite forces are exerted on each other, due to Newton´s 3rd Motion Law).

You mean the CM can be used to track Earth’s motion as if it were a point mass. Indeed, it doesn’t work when the discussion is about more local effects like tidal forces. It also doesn’t work in general. I can have two pairs of all identical mass objects with identical difference in separation by CM, but one is a pair of spheres and the other is a pair of long barbells arranged perpendicularly. The spheres will have more gravitational attraction than the crossed barbells despite the same mass and distanace between their respective centers of mass, because most of the mass is nowhere near that center like it is with the spheres. So point taken. You can’t treat objects as point masses when the difference matters.You describe tidal forces nicely there. Yes, they put stresses on things, enough to tear them apart if they get within the Roche limit.

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Initially I didn´t agree, and insisted: only previously mentioned outward internal forces could be considered “centrifugal”, to fully satisfy Newton´s 2nd Motion Law …

And since nothing is accelerating outward, this would be consistent with an outward centrifugal force not being real. Sorry, just thinking out loud when you haven’t got to the point yet.

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In any case, both ways are equivalent, because deducting required centripetal force vector from moon´s pull on a given earth particle (the force the particle “interchanges” with the rest of the planet), gives same result than adding the “arguably” called centrifugal force vector to moon´s pull …

Still agree… As I said, I suspect the difference is interpretational.

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But it is necessary to tackle the “apparently” insolvable fact that earth´s CM is actually accelerating … It is rather tough to explain through posts here. I did have in mind a couple of ways. With your post, yet another. In chronological order:A) One in relation to the erroneously (to me) used concept of “non-ínertial reference system" …

That seems unnecessary if you get rid of the sun and everything else and just have Earth and moon in the inertial frame of their mutual CM. Keeping the sun and its frame lets you see that the ‘centrifugal’ tide is actually accelerating away from the moon at full moon.

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B) Another way (though kind of equivalent) in relation to the scenarios used by D.C., imagining earth not revolving …

So they wouldn’t be called tides anymore because the distortion would be permanent and not noticed. This is assuming Earth rotating in sync with the month, not stopped altogether, which would have the tides occurring twice a month. The day is lengthening, but it will never be longer than the lunar month.

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C) Last one, in relation to what you say: “… If the two forces (moon´s pull and centrifugal force) where equal and opposite, the particles at earth’s center would be moving in a straight line, but they’re not". ANOTHER DAY I´ll try and do it.

Sorry, but I never figured out from these three ways what this insolvable fact is concerning earth’s CM accelerating. My comment quoted in C there was just another way of pointing out why I don’t consider inertia to be a force.

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Anyhow I´ll copy and paste something I´ve already said a couple of times (way A), needing to elaborate on).PmbPhy had said:"In particular, the centrifugal force is an inertial force which is observed in rotating (non-inertial) frames”.I replied:"I´m afraid that is rather confusing ... What do you mean with "... "which is observed in …"?Does it mean it is a real force "observed" by us? Or, in that "particular" case, is it rather a kind of tricky tool WE apply to bring a "fictitious" situation (an accelerating frame of reference) back to reality (with all REAL effects which have disappeared in that frame of reference)?"

Centrifugal force seems to be the force that holds the water in the pail being swung around, similar to how gravity holds water in the same pail hanging from my arm. But both are wrong. It is the pull of my arm that holds the water in the pail in both cases. Without that, the water would drift out of the pail, even in the gravity field. Water isn’t held with pails on the space station, despite no lack of gravity up there.

#316 HalcWhen I read carefully your post, I´ll try and refute what found necessary ... But just one thing now.Months ago I discussed the issue with a physicist on the email. I even posted here part of the "story".He is in line of your and D.C.´s stand, as far as moon (and sun) related tides are concerned.But, on the site linked below, he explained why we weight less on the equator, and why the so called "equatorial bulge" (the circular and permanent one due to daily earth spinning) happens ...I commented on that here (#141... for some reason I started referring to tidal effects on the moon, due to similar reasons to the ones on our planet):"Moon is rotating around the barycenter, and centripetal (mainly earth pull) and centrifugal forces do happen at different parts of it. And also own moon gravitational forces.At closer moon half, earth attraction prevails. At further moon half, earth pull is insufficient to supply the centripetal force necessary for the rotation ... Own moon gravity has to supply the deficit, and all those further parts kind of lighten, due to same reasons explained on:https://www.lockhaven.edu/~dsimanek/scenario/centrip.htm"The figure shows the force vectors W and mg, which are the only forces acting on the man. The vector F is their sum. W is directed along the radius of the Earth. Being the radial component of the net force (it is the net force in this case), its size is a = v2/R (the centripetal force). Now compare these two cases. On the non-rotating Earth the man's weight was of size mg. Remember, the weight of an object is the force required to support it, i.e., the force exerted upward by the weighing scale. With the Earth rotating, that force is smaller than before. The contact force between the man's feet and the scale is reduced. But all other such stress forces are reduced as well, within the man, within the scale's springs, within the body of the Earth itself. This causes a slight decompression of these materials, a relaxation of the spring in the scales. In fact, the entire body of the earth expands slightly and the man and scale move outward from the axis of rotation slightly, until forces come into balance with the requirements of rotational stability at the new radius. This is the reason for the equatorial bulge of the Earth due to its own axial rotation"Please kindly analyze his explanation, and say if you agree or doesn´t ...He certainly doesn´t use the term "centrifugal force", but the root of his argument is on the REAL effects of INERTIA ... If that happens as an inertial reaction of the water (and the man´s mass) when "forced" to follow the equatorial curved path, why wouldn´t happen something similar when the water has to revolve, following during some 28 days a circumference (considering only earth-moon "dance") of a radius equal to the distance between earth CM and earth-moon barycenter ??

You should perhaps quote the parts of mine that you are refuting, because I pretty much agree with your whole post except the last bit.I looked up centrifugal force and realized it is not a fictional force. In an inertial frame, there is no such force. But it is a term legitimately used in rotating frames, where none of Newton's 3 laws apply. An object at rest tends to accelerate, and the force that causes this acceleration is centrifugal. No reaction stands opposite this action. Moving objects do not follow straight lines. Not sure if all such accelerated motion can be considered centrifugal since the acceleration is more typically inward instead of outward, and 'centrifugal' is typically used only for the tendency towards outward. If not, I don't know the name of the reactionless force that appears to accelerate objects inward in such rotating frames.

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"This is the reason for the equatorial bulge of the Earth due to its own axial rotation[/b]"Please kindly analyze his explanation, and say if you agree or doesn´t ...

This part I is fine. He doesn't really discuss tides, but it is mentioned once in the 'stress' section. Yes, stress causes tides. Such stresses can kill you if it is strong enough.

I took issue with his blocky diagram above in the 'stress' section, which shows two stacked blocks in a gravitational field and a weight force preventing the whole setup from accelerating from the gravity. It labels the spring as tension, despite the arrows depicting the spring pushing on the two blocks, which would compress the spring. The arrows are correct, but text incorrectly labels this as tension. If the arrows represent tension T, they need to point the other way, and the tension would be negative. So while I understand the concept of stress and strain (I took civil engineering courses), the description of it there needs a bit of review.

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He certainly doesn´t use the term "centrifugal force", but the root of his argument is on the REAL effects of INERTIA ...

Yes. In an inertial frame, the equator bulge occurs from a combination of at least 3 effects: You weigh less (and so does the water) there because you're further from the typical particles that make up the planet, and because part of the gravitational force is used for acceleration and need not be resisted by the force of weight from below. The third part is due to the fact that the typical particle is concentrated more below and less to the sides, and this component makes you weigh a bit more, but nowhere near enough to overcome the negative effects of the other two.The second component I listed is the inertial one.

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If that happens as an inertial reaction of the water (and the man´s mass) when "forced" to follow the equatorial curved path, why wouldn´t happen something similar when the water has to revolve, following during some 28 days a circumference (considering only earth-moon "dance") of a radius equal to the distance between earth CM and earth-moon barycenter ??

Inertial effects are indeed involved, but not directly like that. The far tide has far more acceleration about the barycenter (which represents the inertial frame of a two-body model), but both tides are the same magnitude, which they would be even if the moon was further away putting the barycenter above the surface. The direct inertial explanation of the tide would push the water down, not upward, since it is now rotating on the same side of the barycenter. Inertial of the water on the sides would tend to carry it to the far side.

So no, I cannot agree with this wording. In a uniform gravitational field (say at a sufficient distance from the galactic black hole to have a gravitational force on earth similar to that of our moon (2e20N), you have the same inertial forces at work, but undetectable tides. It is the non-uniformity of the gravitational field, not inertia, that causes the stress, and tides which are strain resulting from that stress.